import java.util.Scanner;

class Solution{
    /**
     * 问题：前缀和求解
     * 链接：https://www.nowcoder.com/practice/acead2f4c28c401889915da98ecdc6bf?tpId=230&tqId=2021480&ru=/exam/oj&qru=/ta/dynamic-programming/question-ranking&sourceUrl=%2Fexam%2Foj%3Fpage%3D1%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196
     * 时间复杂度：O(1);
     * 空间复杂度：O(N);
     */
    public static void main1(String[] args)
    {
        Scanner in = new Scanner(System.in);
        //预处理
        int n = in.nextInt(),m = in.nextInt();

        int arr[] = new int[n];
        long dp[] = new long[n];

        for (int i = 0; i < n; i++) arr[i] = in.nextInt();

        dp[0] = arr[0];

        for(int i = 1;i < n;i++) dp[i] = arr[i] + dp[i - 1];

        while(m-- != 0){
            int l = in.nextInt() - 1,r = in.nextInt() - 1;

            if(l == 0) System.out.println(dp[r]);
            else System.out.println(dp[r] - dp[l - 1]);

        }
    }

    /**
     * 问题：二维数组前缀和求解
     * 链接：https://www.nowcoder.com/practice/99eb8040d116414ea3296467ce81cbbc?tpId=230&tqId=2023819&ru=/exam/oj&qru=/ta/dynamic-programming/question-ranking&sourceUrl=%2Fexam%2Foj%3Fpage%3D1%26tab%3D%25E7%25AE%2597%25E6%25B3%2595%25E7%25AF%2587%26topicId%3D196
     * 空间复杂度：O(N);
     * 空间复杂度：O(1);
     */
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int n = in.nextInt(),m = in.nextInt(),q = in.nextInt();

        int[][] arr = new int[n + 1][m+1];

        for(int i = 1;i <= n;i++){
            for(int j = 1;j <=m;j++){
                arr[i][j] = in.nextInt();
            }
        }

        long[][] dp = new long[n+1][m + 1];

        for(int i = 1;i <= n;i++){
            for(int j = 1;j <= m;j++){
                dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + arr[i][j];
            }
        }

        while(q-- != 0){
            int x1 = in.nextInt(),y1 = in.nextInt(),x2 = in.nextInt(),y2 = in.nextInt();
            System.out.println(dp[x2][y2] - dp[x1 - 1][y2] - dp[x2][y1 - 1] + dp[x1 - 1][y1 - 1]);
        }
    }

}
